1.4 Dimensional analysis  (Page 3/6)

Checking equations for dimensional consistency

Strategy

Solution

1. There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn:
   
   $\begin{array}{c}[s]=\text{L}\hfill \\ [vt]=[v]·[t]={\text{LT}}^{\mathrm{-1}}·\text{T}={\text{LT}}^0=\text{L}\hfill \\ [0.5a{t}^2]=[a]·{[t]}^2={\text{LT}}^{\mathrm{-2}}·{\text{T}}^2={\text{LT}}^0=\text{L}\text{.}\hfill \end{array}$
   
   All three terms have the same dimension, so this equation is dimensionally consistent.
2. Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the dimensions of each of the three terms appearing in the equation:
   
   $\begin{array}{c}[s]=\text{L}\hfill \\ [v{t}^2]=[v]·{[t]}^2={\text{LT}}^{\mathrm{-1}}·{\text{T}}^2=\text{LT}\hfill \\ [at]=[a]·[t]={\text{LT}}^{\mathrm{-2}}·\text{T}={\text{LT}}^{\mathrm{-1}}.\hfill \end{array}$
   
   None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense .
3. This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless:
   
   $\begin{array}{c}\left[\frac{a{t}^2}{s}\right]=\frac{[a]·{[t]}^2}{[s]}=\frac{{\text{LT}}^{\mathrm{-2}}·{\text{T}}^2}{\text{L}}=\frac{\text{L}}{\text{L}}=1.\hfill \end{array}$
   
   The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:
   
   $\begin{array}{c}[v]={\text{LT}}^{\mathrm{-1}}\hfill \\ \left[\text{sin}\left(\frac{a{t}^2}{s}\right)\right]=1.\hfill \end{array}$

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”

Significance

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t , we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t :

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t :

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.

Summary

• The dimension of a physical quantity is just an expression of the base quantities from which it is derived.
• All equations expressing physical laws or principles must be dimensionally consistent. This fact can be used as an aid in remembering physical laws, as a way to check whether claimed relationships between physical quantities are possible, and even to derive new physical laws.

Problems